#### Geofleur

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## (V^\mu) = \gamma(\textbf{v},c) ##

and

## (A^\mu) = \gamma \left(\frac{d\gamma}{dt}\textbf{v}+\gamma\frac{d\textbf{v}}{dt},c\frac{d\gamma}{dt} \right)##.

where ## \gamma ## is the Lorentz factor, ## \textbf{v} ## is the ordinary three-velocity (which I will assume is entirely along the x-direction), and ## c ## is the speed of light. In a frame of reference in which the particle is instantaneously at rest, ## \gamma = 1 ##, ## \frac{d\gamma}{dt} = 0 ##, and ## v = 0##, giving

## (V^\mu) = (0,c) ##

and

## (A^\mu) = (\frac{d\textbf{v}}{dt},0) ##.

It ought to be possible to derive these same expressions by starting with a particle moving with velocity ##\textbf{v}## and then Lorentz boosting into its instantaneous rest frame. However, when I try to do this I get the wrong result for the four-acceleration. The matrix for the Lorentz transformation giving primed quantities in terms of unprimed quantities is

## [L^{\mu}_{\nu}] =

\begin{bmatrix}

\gamma & 0 & 0 & -\gamma v / c \\

0 & 1 & 0 & 0 \\

0 & 0 & 1 & 0 \\

-\gamma v / c & 0 & 0 & \gamma \\

\end{bmatrix}

##

The four-acceleration should transform as ## A'^{\mu}=L^{\mu}_{\nu}A^\nu##. But then applying this formula to ##A'^1## I get

## A'^1 = L^1_1 A^1 + L^1_2 A^2 + L^1_3 A^3 + L^1_4 A^4 = \gamma^2 \frac{d\gamma}{dt}v + \gamma^3 \frac{dv}{dt} - \gamma^2 \frac{d\gamma}{dt}v = \gamma^3 \frac{dv}{dt}##, which is different from the correct result by a factor of ##\gamma^3##! Where have I gone wrong?